a, ĐKXĐ : \(\left[{}\begin{matrix}x\le-3\\x\ge0\end{matrix}\right.\)

TH1 : \(x\le-3\) ( LĐ )

TH2 : \(x\ge0\)

BPT \(\Leftrightarrow x^2+2x+x^2+3x+2\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge4x^2\)

\(\Leftrightarrow\sqrt{\left(x^2+2x\right)\left(x^2+3x\right)}\ge x^2-\dfrac{5}{2}x\)

\(\Leftrightarrow2\sqrt{\left(x+2\right)\left(x+3\right)}\ge2x-5\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\x\ge-2\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\4x^2+20x+24\ge4x^2-20x+25\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}0\le x< \dfrac{5}{2}\\x\ge\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow x\ge0\)

Vậy \(S=R/\left(-3;0\right)\)

b.

Với \(x=0\) không phải nghiệm

Với \(x\ne0\) hệ tương đương:

\(\left\{{}\begin{matrix}\dfrac{y}{x^2}+\dfrac{y^2}{x}=-6\\\dfrac{1}{x^3}+y^3=19\end{matrix}\right.\)

Đặt \(\left(\dfrac{1}{x};y\right)=\left(u;v\right)\) ta được: \(\left\{{}\begin{matrix}uv^2+u^2v=-6\\u^3+v^3=19\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3uv^2+3u^2v=-18\\u^3+v^3+19\end{matrix}\right.\)

Cộng vế với vế:

\(\left(u+v\right)^3=1\Rightarrow u+v=1\)

Thay vào \(u^2v+uv^2=-6\Rightarrow uv=-6\)

Theo Viet đảo, u và v là nghiệm của:

\(t^2-t-6=0\) \(\Rightarrow\left[{}\begin{matrix}t=-2\\t=3\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(-2;3\right);\left(3;-2\right)\)

\(\Rightarrow\left(\dfrac{1}{x};y\right)=\left(-2;3\right);\left(3;-2\right)\)

\(\Rightarrow\left(x;y\right)=\left(-\dfrac{1}{2};3\right);\left(\dfrac{1}{3};-2\right)\)

a.

ĐKXĐ: \(x\ne3\)

- Với \(x\ge0\) pt trở thành:

\(\dfrac{x^2-x-12}{x-3}=2x\Rightarrow x^2-x-12=2x^2-6x\)

\(\Leftrightarrow x^2-5x+12=0\) (vô nghiệm)

- Với \(x< 0\) pt trở thành:

\(\dfrac{x^2+x-12}{x-3}=2x\Rightarrow\dfrac{\left(x-3\right)\left(x+4\right)}{x-3}=2x\)

\(\Rightarrow x+4=2x\Rightarrow x=4>0\) (ktm)

Vậy pt đã cho vô nghiệm

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

pt đã cho \(\Leftrightarrow x^2+x+2-\left(2x+3\right)\sqrt{x^2+x+2}+x^2+x-1=-\left(2x+3\right)\)

\(\Leftrightarrow x^2+x+2-\left(2x+3\right)\sqrt{x^2+x+2}+x^2+3x+2=0\)

Đặt \(t=\sqrt{x^2+x+2}\left(t\ge0\right)\)  pt trở thành

\(t^2-\left(2x+3\right)t+x^2+3x+2=0\) (*)

pt (*) có biệt thức \(\Delta=\left(2x+3\right)^2-4\left(x^2+3x+2\right)=1\)

\(t_1=\frac{2x+3+1}{2}=x+2\) \(\Leftrightarrow\begin{cases}x\ge-2\\\sqrt{x^2+x+2}=x+2\end{cases}\Leftrightarrow x=-\frac{2}{3}}\)

 \(t_2=\frac{2x+3-1}{2}=x+1\) 

\(\Leftrightarrow\begin{cases}x\ge-1\\\sqrt{x^2+x+2}=x+1\end{cases}\Leftrightarrow x=1}\)

x=1

x=-0,(6)

ĐKXĐ: \(2x^2-3x+1<>0;2x^2+x+1<>0\)

=>(2x-1)(x-1)<>0 và \(x^2+\frac12x+\frac12<>0\)

=>x∉{1;1/2} và \(x^2+\frac12x+\frac{1}{16}+\frac{7}{16}<>0\)

=>x∉{1;1/2} và \(\left(x+\frac14\right)^2+\frac{7}{16}<>0\) (luôn đúng)

=>x∉{1;1/2}

Ta có: \(\frac{2x}{2x^2-3x+1}-\frac{x}{2x^2+x+1}=\frac{-3}{2}\)

=>\(\frac{2x\left(2x^2+x+1\right)-x\left(2x^2-3x+1\right)}{\left(2x^2-3x+1\right)\left(2x^2+x+1\right)}=\frac{-3}{2}\)

=>\(\frac{4x^3+2x^2+2x-2x^3+3x^2-x}{\left(2x^2-3x+1\right)\left(2x^2+x+1\right)}=\frac{-3}{2}\)

=>\(\frac{2x^3+5x^2+x}{\left(2x^2-3x+1\right)\left(2x^2+x+1\right)}=\frac{-3}{2}\)

=>\(-3\left\lbrack\left(2x^2+1\right)^2-2x\cdot\left(2x^2+1\right)-3x^2\right\rbrack=2\left(2x^3+x+5x^2\right)\)

=>\(-3\left(2x^2+1\right)^2+6x\left(2x^2+1\right)+9x^2-2x\left(2x^2+1\right)-10x^2=0\)

=>\(-3\left(2x^2+1\right)^2+4x\left(2x^2+1\right)-x^2=0\)

=>\(3\left(2x^2+1\right)^2-4x\left(2x^2+1\right)+x^2=0\) (1)

Đặt \(a=2x^2+1\)

(1) sẽ trở thành: \(3a^2-4xa+x^2=0\)

=>\(3a^2-3a\cdot x-a\cdot x+x^2=0\)

=>3a(a-x)-x(a-x)=0

=>(a-x)(3a-x)=0

TH1: a-x=0

=>\(2x^2-x+1=0\)

=>\(x^2-\frac12x+\frac12=0\)

=>\(\left(x-\frac14\right)^2+\frac{7}{16}=0\) (vô lý)

TH2; 3a-x=0

=>\(6x^2+3-x=0\)

=>\(6x^2-x+3=0\)

\(\Delta=\left(-1\right)^2-4\cdot6\cdot3=1-72=-71<0\)

=>x∈∅